Integrand size = 22, antiderivative size = 126 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {(b B-6 A c) (b+2 c x) \sqrt {b x+c x^2}}{8 c}+\frac {(b B-6 A c) \left (b x+c x^2\right )^{3/2}}{3 b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}-\frac {b^2 (b B-6 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{8 c^{3/2}} \]
1/3*(-6*A*c+B*b)*(c*x^2+b*x)^(3/2)/b+2*A*(c*x^2+b*x)^(5/2)/b/x^2-1/8*b^2*( -6*A*c+B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(3/2)+1/8*(-6*A*c+B*b)* (2*c*x+b)*(c*x^2+b*x)^(1/2)/c
Time = 0.18 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {\sqrt {x (b+c x)} \left (\sqrt {c} \left (3 b^2 B+4 c^2 x (3 A+2 B x)+2 b c (15 A+7 B x)\right )+\frac {3 b^2 (b B-6 A c) \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )}{\sqrt {x} \sqrt {b+c x}}\right )}{24 c^{3/2}} \]
(Sqrt[x*(b + c*x)]*(Sqrt[c]*(3*b^2*B + 4*c^2*x*(3*A + 2*B*x) + 2*b*c*(15*A + 7*B*x)) + (3*b^2*(b*B - 6*A*c)*Log[-(Sqrt[c]*Sqrt[x]) + Sqrt[b + c*x]]) /(Sqrt[x]*Sqrt[b + c*x])))/(24*c^(3/2))
Time = 0.27 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1220, 1131, 1087, 1091, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1220 |
\(\displaystyle \frac {(b B-6 A c) \int \frac {\left (c x^2+b x\right )^{3/2}}{x}dx}{b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}\) |
\(\Big \downarrow \) 1131 |
\(\displaystyle \frac {(b B-6 A c) \left (\frac {1}{2} b \int \sqrt {c x^2+b x}dx+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b B-6 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {(b B-6 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(b B-6 A c) \left (\frac {1}{2} b \left (\frac {(b+2 c x) \sqrt {b x+c x^2}}{4 c}-\frac {b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{3/2}}\right )+\frac {1}{3} \left (b x+c x^2\right )^{3/2}\right )}{b}+\frac {2 A \left (b x+c x^2\right )^{5/2}}{b x^2}\) |
(2*A*(b*x + c*x^2)^(5/2))/(b*x^2) + ((b*B - 6*A*c)*((b*x + c*x^2)^(3/2)/3 + (b*(((b + 2*c*x)*Sqrt[b*x + c*x^2])/(4*c) - (b^2*ArcTanh[(Sqrt[c]*x)/Sqr t[b*x + c*x^2]])/(4*c^(3/2))))/2))/b
3.1.84.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x ^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e *f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)) Int[(d + e*x )^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] && !IGtQ[m + p + 1, 0 ]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 ]
Time = 0.14 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {\left (\frac {3}{2} A \,b^{2} c -\frac {1}{4} B \,b^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\left (\frac {5 \left (\frac {7 B x}{15}+A \right ) b \,c^{\frac {3}{2}}}{2}+x \left (\frac {2 B x}{3}+A \right ) c^{\frac {5}{2}}+\frac {B \sqrt {c}\, b^{2}}{4}\right ) \sqrt {x \left (c x +b \right )}}{2 c^{\frac {3}{2}}}\) | \(82\) |
risch | \(\frac {\left (8 B \,c^{2} x^{2}+12 A \,c^{2} x +14 B b c x +30 A b c +3 B \,b^{2}\right ) x \left (c x +b \right )}{24 c \sqrt {x \left (c x +b \right )}}+\frac {b^{2} \left (6 A c -B b \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{16 c^{\frac {3}{2}}}\) | \(97\) |
default | \(B \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )+A \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )\) | \(176\) |
1/2*((3/2*A*b^2*c-1/4*B*b^3)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(5/2*(7/ 15*B*x+A)*b*c^(3/2)+x*(2/3*B*x+A)*c^(5/2)+1/4*B*c^(1/2)*b^2)*(x*(c*x+b))^( 1/2))/c^(3/2)
Time = 0.26 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.63 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\left [-\frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (8 \, B c^{3} x^{2} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{48 \, c^{2}}, \frac {3 \, {\left (B b^{3} - 6 \, A b^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (8 \, B c^{3} x^{2} + 3 \, B b^{2} c + 30 \, A b c^{2} + 2 \, {\left (7 \, B b c^{2} + 6 \, A c^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{24 \, c^{2}}\right ] \]
[-1/48*(3*(B*b^3 - 6*A*b^2*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)* sqrt(c)) - 2*(8*B*c^3*x^2 + 3*B*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^ 3)*x)*sqrt(c*x^2 + b*x))/c^2, 1/24*(3*(B*b^3 - 6*A*b^2*c)*sqrt(-c)*arctan( sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*x^2 + 3*B*b^2*c + 30*A*b*c^2 + 2*(7*B*b*c^2 + 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^2]
\[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\int \frac {\left (x \left (b + c x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{x^{2}}\, dx \]
Time = 0.18 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} B b x - \frac {B b^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{16 \, c^{\frac {3}{2}}} + \frac {3 \, A b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, \sqrt {c}} + \frac {1}{3} \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} B + \frac {3}{4} \, \sqrt {c x^{2} + b x} A b + \frac {\sqrt {c x^{2} + b x} B b^{2}}{8 \, c} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{2 \, x} \]
1/4*sqrt(c*x^2 + b*x)*B*b*x - 1/16*B*b^3*log(2*c*x + b + 2*sqrt(c*x^2 + b* x)*sqrt(c))/c^(3/2) + 3/8*A*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c ))/sqrt(c) + 1/3*(c*x^2 + b*x)^(3/2)*B + 3/4*sqrt(c*x^2 + b*x)*A*b + 1/8*s qrt(c*x^2 + b*x)*B*b^2/c + 1/2*(c*x^2 + b*x)^(3/2)*A/x
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.85 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x} {\left (2 \, {\left (4 \, B c x + \frac {7 \, B b c^{2} + 6 \, A c^{3}}{c^{2}}\right )} x + \frac {3 \, {\left (B b^{2} c + 10 \, A b c^{2}\right )}}{c^{2}}\right )} + \frac {{\left (B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {3}{2}}} \]
1/24*sqrt(c*x^2 + b*x)*(2*(4*B*c*x + (7*B*b*c^2 + 6*A*c^3)/c^2)*x + 3*(B*b ^2*c + 10*A*b*c^2)/c^2) + 1/16*(B*b^3 - 6*A*b^2*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(3/2)
Timed out. \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^2} \, dx=\int \frac {{\left (c\,x^2+b\,x\right )}^{3/2}\,\left (A+B\,x\right )}{x^2} \,d x \]